A small rock is thrown vertically upward with a speed of 27.0 m/s from the edge of the roof of a 21.0-m-tall building. The rock doesn't hit the building on its way back down and lands in the street below. Ignore air resistance. Part A: What is the speed of the rock just before it hits the street?
Part B: How much time elapses from when the rock is thrown until it hits the street?

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yesquitpipe yesquitpipe · Answer 1 · 2019-09-16T22:56:30+02:00

Answer:

A. 33.77 m/s

B. 6.20 s

Explanation:

Frame of reference:

Gravity g=-9.8 m/s^2; Initial position (roof) y=0; Final Position street y= -21 m

Initial velocity upwards v= 27 m/s

Part A. Using kinematics expression for velocities and distance:

[tex]V_{final}^{2}=V_{initial}^{2}+2g(y_{final}-y_{initial})\\V_{f}^{2}=27^{2}-2*9.8(-21-0)=33.77 m/s[/tex]

Part B. Using Kinematics expression for distance, time and initial velocity

[tex]y_{final}=y_{initial}+V_{initial}t+\frac{1}{2} g*t^{2}\\==> -0.5*9.8t^{2}+27t+21=0\\==> t_1 =-0.69 s\\t_2=6.20 s[/tex]

Since it is a second order equation for time, we solved it with a calculator. We pick the positive solution.